The statement `(p ^^q) implies ~p` is a

Options

A. Tautology
B. contradiction
C. Neither tautology nor contradiction
D. None of these

Answer

To determine whether the statement \((p \land q) \implies \neg p\) is a tautology, contradiction, or neither, we can construct a truth table. Here’s the step-by-step solution:

### Step 1: Identify the components of the statement
The statement consists of:
- \(p\): a proposition that can be true (T) or false (F).
- \(q\): another proposition that can also be true (T) or false (F).
- \(\neg p\): the negation of \(p\).
- \(p \land q\): the conjunction (AND) of \(p\) and \(q\).
- \((p \land q) \implies \neg p\): the implication from \(p \land q\) to \(\neg p\).

### Step 2: Create a truth table
We will create a truth table with columns for \(p\), \(q\), \(p \land q\), \(\neg p\), and \((p \land q) \implies \neg p\).

| \(p\) | \(q\) | \(p \land q\) | \(\neg p\) | \((p \land q) \implies \neg p\) |
|-------|-------|----------------|------------|----------------------------------|
| T | T | T | F | F |
| T | F | F | F | T |
| F | T | F | T | T |
| F | F | F | T | T |

### Step 3: Fill in the truth values
1. **For \(p = T\) and \(q = T\)**:
- \(p \land q = T\)
- \(\neg p = F\)
- \((p \land q) \implies \neg p = F\) (since T implies F is F)

2. **For \(p = T\) and \(q = F\)**:
- \(p \land q = F\)
- \(\neg p = F\)
- \((p \land q) \implies \neg p = T\) (since F implies anything is T)

3. **For \(p = F\) and \(q = T\)**:
- \(p \land q = F\)
- \(\neg p = T\)
- \((p \land q) \implies \neg p = T\)

4. **For \(p = F\) and \(q = F\)**:
- \(p \land q = F\)
- \(\neg p = T\)
- \((p \land q) \implies \neg p = T\)

### Step 4: Analyze the results
Now, we look at the final column of the truth table:
- The values are: F, T, T, T.

Since the statement \((p \land q) \implies \neg p\) is not always true (it is false when both \(p\) and \(q\) are true), it is not a tautology.
Since it is not always false (it is true in three cases), it is not a contradiction either.

### Conclusion
The statement \((p \land q) \implies \neg p\) is neither a tautology nor a contradiction.

**Final Answer**: Option 3: Neither tautology nor contradiction.